Probability - Testing for diseases I am just learning probability in my Discrete Structures class and am very lost. This is the example given in the book and I have no idea how to solve this problem. **Problem:** Suppose one in 1000 people have a certain disease. Suppose medical testing is not perfect (as is the case in real life) and consequently, only 99% of the people with the disease are tested positive. Suppose 2% of the people who don’t have the disease also test positive. What is the probability of actually having the disease, given someone tests positive? **Things I do know:** * People with disease = 1/1000 * Tested positive with disease = 99/100 * People tested positive who don't have disease = 2/100 I'm not sure where to got from here with the information that I know. What are the next steps?

As nomen pointed out, you need Bayes' Theorem here but you will find it helpful the expand the denominator as follows

$P(A|B) = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|\bar{A})P(\bar{A})}$

where the bars indicate complements. In particular, let A be having the disease and B be testing positive. Then

$P(B|A) = \frac{99}{100}$

$P(A) = \frac{1}{1000}$

$P(B|\bar{A}) = \frac{2}{100}$

$P(\bar{A}) = 1 - P(A)$

Plug and chug from here.